Question: $\text D = \left[\begin{array}{rr}2 & 0 \\ 4 & 1\end{array}\right]$ and $\text C = \left[\begin{array}{rrr}2 & 4 & 4 \\ 4 & 2 & 4\end{array}\right]$ Let $\text {H = DC}$. Find $\text H$. $ {H = }$
Answer: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{D}$ and the first column of $\text{C}$. $ \text {H}=\left[\begin{array}{rr}{2} & {0} \\ 4 & 1\end{array}\right]\left[\begin{array}{rr} {2} & 4 & 4 \\ {4} & 2 & 4\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(2,0)\cdot(2,4)\\\\ &=2 \cdot 2 + 0\cdot 4\\\\ &=4 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4 \cdot 4 + 1\cdot 2 = 18$ (Choice B) B $2 \cdot 4 + 0\cdot 2 = 8$ (Choice C) C $4 \cdot 2 + 1\cdot 4 = 12$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}= \left[\begin{array}{rrr}4 & 8 & 8 \\ 12 & 18 & 20\end{array}\right] $